|
|
| Line 1: |
Line 1: |
| − | Could you justify at the end of solution II and solution III the chain rule? Why are the derivative of ''q_i'' with respect to ''Q_k'' zero?
| + | Cravens Solution to Problem 9 and 10 |
| − | | + | |
| − | | + | |
| − | Hallo, why is the derivate of P_s with respect to Q_j equal zero? you do this in the last step of solution III.
| + | |
| − | If this is true, then your equation (3), derivate of P_i w.r.t. p_k, would be equal zero too.
| + | |
| − | maybe it could be useful in solution II and III, that eq. (3) is zero...
| + | |
| − | | + | |
| − | | + | |
| − | | + | |
| − | '''Alternative solution:'''
| + | |
| − | | + | |
| − | Here's a proof I've found on my usual source of knowledge (meaning a youtube-lecture):
| + | |
| − | | + | |
| − | '''Claim:''' For a generating function of type one (that's the smartman's name for the Phi in our problem) it holds: \( \{ Q, P \} = 1 \).
| + | |
| − | | + | |
| − | ''Proof:''
| + | |
| − | | + | |
| − | We first fix the \( q \) coordinates in the momentum equation and derive by \( p \), denoted: \( \Phi_p |_q\).
| + | |
| − | | + | |
| − | So we get:
| + | |
| − | | + | |
| − | '''Equation 1:'''
| + | |
| − | | + | |
| − | $$ 1 = \Phi_{q Q} Q_p |_q \Rightarrow Q_p |_q = \frac{1}{\Phi_{q Q}} $$
| + | |
| − | | + | |
| − | '''Equation 2:'''
| + | |
| − | | + | |
| − | Holding \( p \) fixed and derive by \( q \):
| + | |
| − | | + | |
| − | $$ 0 = \frac{\partial p}{\partial q} = \Phi_{qq} + \Phi_{qQ}Q_{q}|_p \Rightarrow Q_{q}|_p = - \frac{\Phi_{qq}}{\Phi_{qQ}} $$
| + | |
| − | | + | |
| − | '''Equation 3:'''
| + | |
| − | | + | |
| − | Holding \( q \) fixed and derive by \( p \):
| + | |
| − | | + | |
| − | $$ P_p |_q = - \Phi_{QQ}Q_q |_p \Rightarrow P_p |_q = - \frac{\Phi_{QQ}}{\Phi_{qQ}} $$
| + | |
| − | | + | |
| − | where we used equation 1 again.
| + | |
| − | | + | |
| − | '''Equation 4:'''
| + | |
| − | | + | |
| − | Holding \( p \) fixed and derive by \( q \):
| + | |
| − | | + | |
| − | $$ P_q |_p = - \Phi_{Qq} - \Phi_{QQ} Q_q |_p \Rightarrow P_q |_p = - \Phi_{Qq} + \frac{\Phi_{QQ}\Phi_{qq}}{\Phi_{qQ}} $$
| + | |
| − | | + | |
| − | using equation 2.
| + | |
| − | | + | |
| − | We are now prepared to calculate:
| + | |
| − | | + | |
| − | $$ \{ Q, P \} = Q_qP_p - Q_pP_q = \frac{\Phi_{QQ}}{\Phi_{qQ}} \frac{\Phi_{qq}}{\Phi_{qQ}} - \frac{1}{\Phi_{q Q}} \left( - \Phi_{Qq} + \frac{\Phi_{QQ}\Phi_{qq}}{\Phi_{qQ}} \right) = 1 $$ by just cancelling out the terms. \( \square \)
| + | |
| − | | + | |
| − | ( ''My first idea would have been:''
| + | |
| − | | + | |
| − | By the script:
| + | |
| − | | + | |
| − | $$ \{ Q, P \} = \langle \nabla Q , J \nabla P \rangle_{\mathbb{R}^{2n}} = 1 $$
| + | |
| − | | + | |
| − | with \( \nabla = \left( \frac{\partial}{\partial q_1}, ..., \frac{\partial}{\partial p_n} \right) \) and
| + | |
| − | | + | |
| − | $$ J = \begin{bmatrix}
| + | |
| − | 0 & I_n \\
| + | |
| − | -I_n & 0 \\
| + | |
| − | \end{bmatrix} $$
| + | |
| − | | + | |
| − | and this should result in the things we want to show (by some symplectic argument or so...).)
| + | |
| − | | + | |
| − | ''But it might be better to just show this (the property is surely right, found it in a classical mechanics textbook (''guess that would be [http://www.scielo.org.mx/pdf/rmfe/v57n2/v57n2a4.pdf here], p2 top right -Nik'')):''
| + | |
| − | | + | |
| − | | + | |
| − | \( \vdash : \{ f, g \}_{p, q} = \{ P, Q \} \{ f, g \}_{P, Q} \)
| + | |
| − | | + | |
| − | ''Proof''
| + | |
| − | | + | |
| − | With simplified notation:
| + | |
| − | | + | |
| − | \( \{ f, g \}_{p, q} = \sum_k \left( \frac{\partial f}{\partial q_k} \frac{\partial g}{\partial p_k} - \frac{\partial f}{\partial p_k} \frac{\partial g}{\partial q_k} \right) = \sum_k \left( \frac{\partial f}{\partial Q} \frac{\partial Q}{\partial q_k} + \frac{\partial f}{\partial P} \frac{\partial P}{\partial q_k} \right) \left( \frac{\partial g}{\partial Q} \frac{\partial Q}{\partial p_k} + \frac{\partial g}{\partial P} \frac{\partial P}{\partial p_k} \right) - \sum_k \left( \frac{\partial f}{\partial Q} \frac{\partial Q}{\partial p_k} + \frac{\partial f}{\partial P} \frac{\partial P}{\partial p_k} \right) \) \(\left( \frac{\partial g}{\partial Q} \frac{\partial Q}{\partial q_k} + \frac{\partial g}{\partial P} \frac{\partial P}{\partial q_k} \right) \)
| + | |
| − | | + | |
| − | \( = \sum_k \left( \frac{\partial Q}{\partial q_k} \frac{\partial P}{\partial p_k} - \frac{\partial Q}{\partial p_k} \frac{\partial P}{\partial q_k} \right) \left( \frac{\partial f}{\partial Q} \frac{\partial g}{\partial P} - \frac{\partial f}{\partial P} \frac{\partial g}{\partial Q} \right) = \{ Q, P \}_{q, p} \cdot \{ f, g \}_{Q, P} \) \( \square \)
| + | |
| − | | + | |
| − | | + | |
| − | | + | |
| − | Now compute: \( \{ Q, P \}_{q, p} \cdot \{ Q_i, Q_j \}_{Q, P} = \{ Q, P \}_{q, p} \cdot \{ P_i, P_j \}_{Q, P} = 0 \) which follows from the properties of the Poisson-bracket.
| + | |
| − | | + | |
| − | Furthermore: \( \{ Q, P \}_{q, p} \cdot \{ Q_i, P_j \}_{Q, P} = \{ Q, P \}_{q, p} \cdot \delta_{i j} = \delta_{i j} \)
| + | |
| − | | + | |
| − | which solves the problem.
| + | |
| − | | + | |
| − | Cheerio, A.
| + | |
| − | | + | |
| − | Ich hab grade alles nochmal durchgerechnet mit mit den festgehalten Indizes und dann funktioniert die kettenregel, bei solution II, weil q dann festgehalten ist. alles andere ist unbeeinflusst also genauso wie in der aktuellen Lösung. man muss einfach am Anfang, wenn man nach q_i ableitet p festhalten und wenn man nach p_i ableitet q festhalten, und dann immer die festgehalten Indizien mitschleppen. Ich hatte jetzt noch keine zeit darüber nachzudenken ob das sinn macht, vielleicht durchschaut das jemand von euch ;) -Carl
| + | |
| − | | + | |
| − | Ladies and gentlemen,
| + | |
| − | a good friend of mine found a very simple modification for problem 7. To justify why the sum in ii) equals the kronecker delta we simply have to take a look at the matrices we defined above. As A*B=1 A and B comute and then it follows directly, so it's pretty immidiate. As for iii) things get easier too as we can then put it in again in the next to last line. So we dont have to assume that dP/dQ equals 0.
| + | |
| − | | + | |
| − | Cheers,
| + | |
| − | | + | |
| − | your [[User:Milfhunter|Milfhunter]] ([[User talk:Milfhunter|talk]]) 15:18, 19 January 2015 (CET)
| + | |
